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Juliette [100K]
2 years ago
7

The relation R is shown below as a list of ordered pairs. R = { (1, 4), (1, 3), (-1, 3), (2, 15) } Which ordered pairs prevent t

his relation from being a function? (1, 4) and (1, 3), because they have the same x-value (1, 3) and (–1, 3), because they have the same y-value
Mathematics
2 answers:
s344n2d4d5 [400]2 years ago
8 0

Answer:

(1,4) and (1,3)

Step-by-step explanation:

A function may not have two y-values assigned to the same x-value, it may have two x-values assigned to the same y-value

Therefore the order pair (1,4) and (1,3) prevents this relation from being a function.

Each x has only one y-value.

Sindrei [870]2 years ago
8 0

Answer: Points (ordered pairs) in a function are allowed to have the same y-value. If some do, they prevent the relation from being one-to-one, but do not prevent the relation from being a function.


(1,4) and (1,3) prevent the relation from being a function.


After you remove one of them, pairs (1,3) and (-1,3) prevent the function from being one to one, which means it has no inverse, but it is a function.


Step-by-step explanation:


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On a cruise of 1125 passengers. 450 were female what fraction of passengers were male?
WINSTONCH [101]

Step-by-step explanation:

\\1125-450=675

\frac{675}{1125}=\frac{3}{5}

4 0
1 year ago
Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |
Bas_tet [7]

Answer:

M=168k

(\bar{x},\bar{y})=(5,\frac{85}{28})

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

M=\int\int \rho dA

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx

Let's solve this integral:

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx      

M=k\int^{9}_{1}21dx

M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}

So the mass will be:

M=21k*8=168k

Now we need to find the x-coordinate of the center of mass.

\bar{x}=\frac{1}{M}\int\int x*\rho dydx

\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx

\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx

\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}

\bar{x}=\frac{21}{168}*40=5

Now we need to find the y-coordinate of the center of mass.

\bar{y}=\frac{1}{M}\int\int y*\rho dydx

\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx

\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx

\bar{y}=\frac{255}{672}\int^{9}_{1}dx

\bar{y}=\frac{255}{672}8=\frac{2040}{672}

\bar{y}=\frac{85}{28}

Therefore the center of mass is:

(\bar{x},\bar{y})=(5,\frac{85}{28})

I hope it helps you!

3 0
2 years ago
El producto de dos números consecutivos es 552 cuáles son esos números?
lara31 [8.8K]

Answer: 23 y 24 ( ó -23 y -24)

Step-by-step explanation:

Dos números consecutivos se escriben como:

n y (n + 1)

done n es un numero entero.

Entonces "El producto de dos números consecutivos es 552"

Se escribe como:

n*(n + 1) = 552

n^2 + n = 552

n^2 + n - 552 = 0

Tenemos una cuadrática, las posibles soluciones son obtenidas con la formula de Bhaskara.

n = \frac{-1 +- \sqrt{1^2 - 4*1*(-552)} }{2} = \frac{-1 +- 47}{2}

Las dos soluciones son.

n = (-1 - 47)/2 = -48/2 = -24

n = (-1 + 47)/2 = 46/2 = 23

Si tomamos la primer solución, n = -24

Entonces los dos números consecutivos son:

n = -24

(n + 1) = -23

Si n = 23 entonces

n + 1 = 24

Lo cual tiene sentido, por que lo único que cambia son los signos, los cuales se cancelarían en la multiplicación.

7 0
1 year ago
On a coordinate plane, a parallelogram has points A (negative 3, 4), B (3, 4), C (1, negative 2), and D (negative 5, negative 2)
vladimir2022 [97]

Answer:

(1,-12)

Step-by-step explanation:

d starts at (-5,-2) if you add 6 to -5 and minus -2 by 10 you will end up with (1,-12)

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1 year ago
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MatroZZZ [7]
The area of the park is 1465m squared, because you would find the area of the square first which is 100m squared and then you would add that to the area of the trapezoid which is 1365m squared and after you add them, you would get 1465m squared. I’m still not sure if this is the exact answer but please check it over to see if I did it correct :)
4 0
2 years ago
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